In response to the increasing weight of airline passengers, the Federal Aviation

Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 190.1 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 36.5 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 22 passengers.

What is the approximate probability (±0.0001) that the total weight of the passengers exceeds 4572 pounds?

Explanation / Answer

This is like asking the probability of the mean exceeding 4572/22 = 207.8181818.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    207.8181818      
u = mean =    190.1      
n = sample size =    22      
s = standard deviation =    36.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.276866828      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.276866828   ) =    0.01139709 [ANSWER]

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